x=t^2-t+2 , y=t^3-3t Find all points (if any) of horizontal and vertical tangency to the curve.

Parametric curve (x(t),y(t)) has a horizontal tangent when its slope dy/dx is zero, i.e. dy/dt=0 and dx/dt!=0 .
Curve has a vertical tangent if its slope approaches infinity i.e. dx/dt=0 and dy/dt!=0
Given equations of the parametric curve are:
x=t^2-t+2
y=t^3-3t
dx/dt=2t-1
dy/dt=3t^2-3
For horizontal tangents:
dy/dt=0
3t^2-3=0
=>3t^2=3
=>t^2=1
=>t=+-1
Corresponding points on the curve can be found by plugging the values of t in the parametric equation,
For t=1,
x_1=1^2-1+2=2 
y_1=1^3-3(1)=-2
For t=-1,
x_2=2^2-2+2=4
y_2=2^3-3(2)=2
Horizontal tangents are at the points (2,-2) and (4,2)
For vertical tangents,
dx/dt=0
2t-1=0
=>t=1/2
Corresponding points on the curve for t=1/2 are,
x=(1/2)^2-1/2+2
x=1/4-1/2+2
x=(1-2+8)/4
x=7/4
y=(1/2)^3-3(1/2)
y=1/8-3/2
y=(1-12)/8
y=-11/8
Vertical tangent is at the point (7/4,-11/8)